Answer:
The air heats up when being compressed and transefers heat to the barrel.
Explanation:
When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.
In an adiabatic transformation:
[tex]P^{1-k} * T^k = constant[/tex]
For air k = 1.4
SO
[tex]P0^{-0.4} * T0^{1.4} = P1^{-0.4} * T1^{1.4}[/tex]
[tex]T1^{1.4} = \frac{P1^{0.4} * T0^{1.4}}{P0^{0.4}}[/tex]
[tex]T1^{1.4} = \frac{P1}{P0}^{0.4} * T0^{1.4}[/tex]
[tex]T1 = T0 * \frac{P1}{P0}^{0.4/1.4}[/tex]
[tex]T1 = T0 * \frac{P1}{P0}^{0.28}[/tex]
SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.
After it was compressed the hot air will exchange heat with the barrel heating it up.
A rectangular plate casting has dimensions 200mm x 100mm x 20mm. The riser for this sand casting mold is in the shape of a sphere. The casting takes 3.5mins to solidify. Calculate the diameter of the riser so that it takes 20% longer for the riser to solidify
Answer:
Diameter of riser =6.02 mm
Explanation:
Given that
Dimensions of rectangular plate is 200mm x 100mm x 20mm.
Volume of rectangle V= 200 x 100 x 20 [tex]mm^3[/tex]
Surface area of rectangle A
A=2(200 x 100+100 x 20 +20 x 200)[tex]mm^2[/tex]
So V/A=7.69
We know that
Solidification times given as
[tex]t=K\left(\dfrac{V}{A}\right)^2[/tex] -----1
Lets take diameter of riser is d
Given that riser is in spherical shape so V/A=d/6
And
Time for solidification of rectangle is 3.5 min then time for solidificartion of riser is 4.2 min.
Lets take [tex]\dfrac{V}{A}=M[/tex]
[tex]\dfrac{M_{rac}}{M_{riser}}=\dfrac{7.69}{\dfrac{d}{6}}[/tex]
Now from equation 1
[tex]\dfrac{3.5}{4.2}=\left(\dfrac{7.69}{\dfrac{d}{6}}\right)^2[/tex]
So by solving this d=6.02 mm
So the diameter of riser is 6.02 mm.
Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters at 400 kPa and 300 K. Stream 3 leaves the control volume at 150 kPa and 270 K. The control volume does 3 kW of work on the surroundings while losing 5 kW of heat. Find the mass flow rate of stream 2. Neglect changes in kinetic and potential energy.
Answer:
0.08kg/s
Explanation:
For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.
The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.
finally you use the two previous equations to make a system and find the mass flows
I attached procedure
Fluid flowing through a constriction or partially opened valve in a pipe system may develop localized low pressures due to increased velocity. If a pipe contains water at 210 °F, what is the minimum absolute pressure that the water can be reduced to and not have cavitation occur?
Answer:
P=41.84psi
Explanation:
Cavitation is a phenomenon that arises when the pressure of a liquid fluid reaches the vapor pressure, this means that if at any point in a system of pipes and pumps the water pressure is equal to or less than the vapor pressure it begins to evaporate, producing explosive bubbles that can cause damage and noise in the system.
Therefore, for this problem, all we have to do is find the water vapor pressure at a temperature of 270F, using thermodynamics tables
Pv@270F=41.84psi
A certain amount of fuel contains 15x102 Btu of energy, and converted into electric energy in a power station having a 12 percent overall efficiency. The average daily (24 hours) demand on the station is 5MW. In how many days will the fuel be totally consumed?
The question involves calculating the duration fuel will last in a power station with 12% efficiency given its energy content and the daily energy demand. A misunderstanding in the conversion of units and efficiency calculation was identified in the attempt, highlighting the need for a corrected mathematical approach involving unit conversions and understanding power station efficiency.
Explanation:A certain amount of fuel contains 15×102 Btu of energy and is converted into electric energy in a power station having a 12 percent overall efficiency. The average daily demand on the station is 5MW. To find out in how many days the fuel will be totally consumed, first, it's essential to calculate the total energy converted into electrical energy considering the efficiency of the power station. Given the power station efficiency, only 12% of the fuel energy is converted into electrical energy. The total energy available for conversion is 15×102 Btu.
First, convert 15×102 Btu into megajoules (MJ) as 1 Btu is approximately equal to 0.00105506 MJ. Therefore, the energy in MJ is 15×102 × 0.00105506 = 1.58259 MJ. Given that 12% of this energy is usable for electricity, the usable energy is 1.58259 MJ × 12% = 0.189911 MJ. Since 1 MW is 1 MJ/sec, and given the station's demand is 5MW over 24 hours, the total consumption is 5MW × 24h = 120 MW/day.
Then, to find out how many days the fuel would last, divide the total usable electrical energy by the daily consumption. However, an error in the calculation process prevents us from completing the result as intended initially. This calculation seems to involve a misunderstanding regarding the conversion of units and the calculation of efficiency. Therefore, a corrected approach should involve accurately converting the initial Btu to MJ or kWh, taking into account the efficiency correctly, and then calculating the duration of fuel consumption based on the daily energy output and demand.
A closed, rigid, 0.50 m^3 tank is filled with 12 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the pressure is P2 = 4 bar. Determine the initial quality, X1, and the heat transfer, in kJ.
Answer:
X1= 41%
heat transfer = -3450.676 KJ
Explanation:
To get the properties for pure substance in a system we need to know at least to properties. These are usually pressure and temperature because they’re easy to measure. In this case we know the initial pressure (20 bar) which is not enough to get all the properties, but they ask to determine quality, this a property that just have meaning in the two-phase region (equilibrium) so with this information we can get the temperature of the system and all its properties.
There is another property that we can calculate from the data. This is the specific volume. This is defined as [tex]\frac{volume}{mass}[/tex]. We know the mass (12 Kg) and we can assume the volume is the volume of the tank (0.5 [tex]m^{3}[/tex]) because they say that the tank was filled.
With this we get a specific volume of
Specific volume = [tex]\frac{0,5 m^{3}}{ 12 kg}= 0.04166667 \frac{m^{3}}{Kg}[/tex]
From the thermodynamic tables we can get the data for the saturated region with a pressure of 20 bar.
Temperature of saturation = 212.385 °C
Specific volume for the saturated steam (vg) = 0.0995805 [tex]\frac{m^{3}}/{Kg}[\tex]
Specific volume for the saturated liquid (vf)= 0.00117675 [tex]\frac{m^{3}}/{Kg}[\tex]
The specific volume that we calculate before 0.04166667 m^3/Kg is between 0.00117675 m^3/Kg and 0.0995805 m^3/Kg so we can be sure that we are in two-phase region (equilibrium).
The quality (X) is defined as the percentage in mass of saturated steam in a mix (Two-phase region)
The relation between specific volume and quality is
[tex]v = (1-x)*v_{f} + x*v_{g}[\tex]
where
v in the specific volume in the condition (0.04166667 m^3/Kg)
vf = Specific volume for the saturated liquid (0.00117675 m^3/Kg)
vg = Specific volume for the saturated steam (0.0995805 m^3/Kg)
x = quality
clearing the equation we get:
[tex]X = \frac{(v-v_{f})}{(v_{g}-v_{f})}[/tex]
[tex]X =\frac{(0.04166667- 0.00117675)}{ (0.0995805 – 0.00117675)} = 0.411[/tex]
The quality is 41%
To calculate the heat transfer we use the next equation.
Q = m * Cp * delta T
Where
Q = heat transfer (Joules, J)
m= mass of the substance (g)
Cp = specific heat (J/g*K) from tables
Delta T = change in temperature in K for this equation.
The mass of the substance is 12 kg or 12000 g for this equation
Cp from tables is 4,1813 J/g*K. You can find this value for water in different states. Here we are using the value for liquid water.
For delta T, we know the initial temperature 212.385 °C.
We also know that the system was cooled. Since we don’t have more information, we can assume that the system was cooled until a condition where all the steam condensates so now we have a saturated liquid. Since we know the pressure (4 bar), we can get the temperature of saturation for this condition from the thermodynamics tables. This is 143.613 °C, so this is the final temperature for the system.
T(K) = T°C +273
T1(K) = 212.385 + 273.15 = 485.535 K
T2 (K) = 143.613 +273.15= 416.763 K
Delta T (K) = (T2-T1) =416.763 K - 485.535 K = -68.772 K
Now we can calculate Q
Q = 12000g * 4,1813 J/g*K* (-68.772 K) = -3450676.36 J or -3450.676 KJ
Is negative because the heat is transfer from the water to the surroundings
A disk is rotating around an axis located at its center. The angular velocity is 0.6 rad/s and the angular acceleration is 0.3 rad/s^2. The radius of the disk is 0.2 m. What is the magnitude of the acceleration at a point located on the outer edge of the disk, in units of m/s?
Answer:
[tex]a=0.0937 \ m/s^2[/tex]
Explanation:
Given that
Angular velocity (ω)= 0.6 rad/s
Angular acceleration (α)= 0.3 [tex]rad/s^2[/tex]
Radius (r)= 0.2 m
We know that is disc is rotating and having angular acceleration then it will have two acceleration .one is radial acceleration and other one is tangential acceleration.
So
[tex]Radial\ acceleration(a_r)=\omega ^2r\ m/s^2[/tex]
[tex]Radial\ acceleration(a_r)=0.6^2 \times0.2 \ m/s^2[/tex]
[tex](a_r)=0.072 \ m/s^2[/tex]
[tex]Tangential\ acceleration(a_t)=\alpha r\ m/s^2[/tex]
[tex]Tangential\ acceleration(a_t)=0.3\times 0.2\ m/s^2[/tex]
[tex](a_t)=0.06 \ m/s^2[/tex]
So the total acceleration ,a
[tex]a=\sqrt{a_t^2+a_r^2}\ m/s^2[/tex]
[tex]a=\sqrt{0.06^2+0.072^2}\ m/s^2[/tex]
[tex]a=0.0937 \ m/s^2[/tex]
Which of the two materials (brittle vs. ductile) usually obtains the largest modulus of toughness and why?
Answer:
The modulus of toughness is greater for ductile material.
Explanation:
Modulus of toughness is defined as the amount of strain energy that a material stores per unit volume. It is equal to the area under stress-strain curve of the material up to the point of fracture.
As we know that the area under the stress-strain curve of a ductile material is much more than the area under the stress strain curve of a brittle material as brittle materials fail at a lesser load hence we conclude that the modulus of toughness is greater for ductile material than a brittle material.
In our experience we can also relate that for same volume a substance such as glass sheet (brittle) fails at lower load as compared to a sheet of steel (ductile) with identical dimensions.
Express the dimensions of (a) force, (b) pressure, (c) energy, and (d) power in terms of primary dimensions.
Answer:
(A) Dimension of force [tex]MLT^{-2}[/tex]
(b) Dimension of energy [tex]ML^2T^{-2}[/tex]
(c) Dimension of power = [tex]ML^2T^{-3}[/tex]
Explanation:
We have to dimension
(a) Force
We know that force F = ma
Dimension of mass is M, and dimension of a is [tex]LT^{-2}[/tex]
So dimension of force [tex]MLT^{-2}[/tex]
(B) Energy
We know that energy = force × displacement
In previous part we can see dimension of force is [tex]MLT^{-2}[/tex]
and dimension of displacement = L
So dimension of energy [tex]ML^2T^{-2}[/tex]
(c) Power
We know that power [tex]p=\frac{energy}{time }[/tex]
In previous part we can see dimension of energy is [tex]ML^2T^{-2}[/tex]
And dimension of time is T
So dimension of power = [tex]ML^2T^{-3}[/tex]
A person, 175 lbm, wants to fly (hoover) on a 4 lbm skateboard of size 2 ft by 0.8 ft. How large a gauge pressure under the board is needed?
Answer:
[tex]p = 15260.643 \ lbf/ft^2[/tex]
Explanation:
person weight is 175 lbm
weight of stake board 4lbm
size of stakeboard = 2ft by 0.8 ft
area of stakeboard is [tex]2*0.8 ft^2 = 1.6 ft^2[/tex]
gauge pressure is given as
[tex]p =\frac{ w_p+w_s}{A} g[/tex]
where is g is acceleration due to gravity = 32.17 ft/sec^2
puttng all value to get pressure value
[tex]= \frac{175 +4}{1.6}* 32.17[/tex]
[tex]p = 15260.643 \ lbf/ft^2[/tex]
Radioactivity of C-14 is used for dating of ancient artifacts. Archeologist determined that 20% of initial amount of C-14 has remained. Estimate the age of this artifact.
Answer:
13282.3 years
Explanation:
The C-14 decays exponentially:
[tex]\frac{dN}{dt} =λ*N[/tex]
The solution for this equation is
[tex]N= N_{o}*e^{- λt}[/tex]
Where:
No = atom number of C-14 in t=0
N = atom number of C-14 now
I= radioactive decay constant
clearing t this equation we get:
[tex]t=-\frac{1}{ λ}*ln\frac{N}{N_{o}}[/tex]
The term 1/I is called half-life and the value for C-14 is 8252 years.
N for this exercise is 0.2No
[tex] t= -8033 * ln \frac{0.2N_{o} }{N_{o}}[/tex]
t = 13282.3 years
Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Consider a drag of 12 kN on the shuttle. Calculate the work done by the shuttle engine.
Answer:
work done = 48.88 × [tex]10^{9}[/tex] J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × [tex]10^{9}[/tex] J
An ideal Otto Cycle with air as the working fluid has a compression ratio of10. The minimum temperature and pressure in the cycle are 25 degrees celsius and 100 kPa respectively. The highest temperature in the cycle is 1000 degrees celsius. Using the PG model for air with properties: k =1.4, c_p = 1 kJ.kg.K, c_v = 0.717 kJ/kg.K, determine the exhaust temperature (temperature at the end of the expansion stroke) in Kelvins.
Answer:
[tex]T_4[/tex]= 506.79 K
Explanation:
Given that
Minimum temperature [tex]T_1[/tex]= 25 C
Pressure = 100 KPa
Highest temperature [tex]T_3[/tex]= 1000 C
Compression ratio r= 10
We know that for ideal Otto cycle
[tex]\dfrac{T_3}{T_4}=r^{\gamma -1}[/tex]
Where [tex]T_3[/tex] is the exhaust temperature.
Now by putting the values
[tex]\dfrac{T_3}{T_4}=r^{\gamma -1}[/tex]
[tex]\dfrac{1000+273}{T_4}=10^{1.4-1}[/tex]
[tex]T_4[/tex]= 506.79 K
A large fraction of the thermal energy generated in the engine of a car is rejected to the air by the radiator through the circulating water. Should the radiator be analyzed as an open system?
Answer:
Yes
Explanation:
Type of control system
1.Open system
When mass and energy is allowed to pass through the system ,then this type of system is called open system.
2.Close system
When only energy is allowed to pass through the system ,then this type of system is called close system.
3.Isolated system
When mass and energy did not allowed to pass through the system ,then this type of system is called isolated system.
In the engine radiator ,mass of hot fluid enters and interact with the cold air then after heat transfer take place between these two fluids,that is why we can say that radiator is an open system.
It become necessity to use the radiator in all type of engine because it produce cooling effect to the engine cylinder .If temperature of engine cylinder will become too high then it will leads to the high thermal stresses and may be it will break the engine cylinder.
Yes, the radiator should be analyzed as an open system.
In thermodynamics, a system is classified as open if both mass and energy can cross its boundaries. The radiator in a car's engine cooling system is designed to dissipate heat from the engine to the air. This dissipation occurs through the circulation of water (or coolant) which absorbs heat from the engine and releases it to the air flowing through the radiator.
In the radiator:
1. Mass Flow: The coolant continuously enters and exits the radiator, carrying thermal energy with it.
2. Energy Transfer: Heat is transferred from the hot coolant to the cooler ambient air through the radiator fins, driven by forced convection due to airflow.
3. Open System Characteristics: Both mass (coolant) and energy (heat) cross the boundaries of the radiator.
Considering these aspects, the radiator facilitates the flow of coolant (mass) and the transfer of thermal energy (heat), making it an open system.
The radiator in a car should be analyzed as an open system due to the continuous flow of coolant and heat transfer involved in its operation. This perspective is essential for accurately understanding and optimizing the cooling process.
The temperature at the outlet of the turbocharger turbine is 260°C, with an exhaust flow rate of 2 kg/min. Estimate the drop in temperature across the turbine, given that the turbine output power is 1kW. Assume exhaust gas to have a specific heat of 1.05 kJ/kg.K, and an ambient temperature of 25°C
Answer:
[tex]\Delta T = 28.57°C[/tex]
Explanation:
given data:
temperature at outlet of turbine = 260°C
Flow rate = 2 kg/min = 0.033 kg/s
output power = 1 kW
specific heat = 1.05 kJ/kg.K
We know that power generated by turbine is equal to change in enthalpy
[tex]W = h_2 - h_1[/tex]
[tex]= mCp(\Delta T)[/tex]
[tex]1*10^3 = 0.0333*1.05*10^3 * \Delta T[/tex]
[tex]\Delta T = \frac{10^3}{0.033*1.05*10^3}[/tex]
[tex]\Delta T = 28.57°C[/tex]
What is 100 kPa in psia?
Answer:
The pressure in pounds per square inch (psi) equals 14.50 psi.
Explanation:
Since psi stands for 'pounds per square inch' and Pascals stands for 'Newtons per square meters'
We can write the given pressure as
[tex]Pressure=100kPa\\\\Pressure=100\times 10^{3}N/m^{2}[/tex]
Now since 1 Newton of force equals 0.22481 Pound force also
since 1 meter equals 39.37 inches
Using the above values we get
[tex]Pressure=100\times 10^{3}\times \frac{0.22481lb}{(39.37)^{2}inch^{2}}[/tex]
thus the pressure becomes
[tex]Pressure=100\times 10^{3}\times \frac{0.22481}{(39.31)^{2}}pounds/inch^{2}\\\\Pressure=14.50psi[/tex]
Q1. In electronic circuits it is not unusual to encounter currents in the microampere range. Assume a 35 μA current, due to the flow of electrons. What is the average number of electrons per second that flow past a fixed reference cross section that is perpendicular to the direction of flow? (5 Points)
Answer:
2.9*10^14 electrons
Explanation:
An Ampere is the flow of one Coulomb per second, so 35 μA = is 35*10^-6 C per second.
An electron has a charge of 1.6*10^-19 C.
35*10^-6 / 1.6*10^-19 = 2.9*10^14 electrons
So, with a current o 35 μA you have an aevrage of 2.9*10^14 electrons flowing past a fixed reference cross section perpendicular to the direction of flow.
One horsepower is the ability to lift 550 pounds a height of one foot in one second or one horsepower is equal to 550ft-lb/s. How many ft-lb/s are there in 32 horsepower?
Answer:
32 horsepower will be equal to 17600 ft-lb/sec
Explanation:
We have given 1 horsepower = 550 ft-lb/sec
We have to convert 32 horsepower into ft-lb/sec
As we know that 1 horsepower is equal to 550 ft-lb/sec
So for converting horsepower to ft-lb/sec we have multiply with 550
We have given 32 horse power
So [tex]32\ horsepower =32\times 550=17600ft-lb/sec[/tex]
So 32 horsepower will be equal to 17600 ft-lb/sec
Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 mm, determine the minimum flow rate that the pump must provide.
Answer:
minimum flow rate provided by pump is 0.02513 m^3/s
Explanation:
Given data:
Exit velocity of nozzle = 20m/s
Exit diameter = 40 mm
We know that flow rate Q is given as
[tex]Q = A \times V[/tex]
where A is Area
[tex]A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2[/tex]
[tex]Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s[/tex]
minimum flow rate provided by pump is 0.02513 m^3/s
What is the primary structural difference between cantilever wing and semi-cantilever wing?
Answer:
The main component of fixed wing aircraft
1.Wings
2.Fuselages
3.Landing gear
4.Stabilizers
5.Flight control surface
Cantilever wing:
A cantilever wing is directly attached to the fuselages and do not have any external support.
Semi cantilever wing:
A cantilever wing does not directly attached to the fuselages and have any external support.The external support may be one two.
Diameters of small commercially available steel rods vary by sixteenths of an inch. Select the required commercial size of the rod required to support a tensile load of 35,000 lb if the tensile stress cannot exceed 20,000 psi.
Answer:
[tex]1\frac{1}{2}[/tex] in dia rod
Explanation:
Given;
Maximum tensile stress = 20,000 psi
Applied tensile load = 35,000 lb
Now,
the area of the steel rod required = [tex]\frac{\textup{Applied tensile load}}{\textup{Maximum tensile stress}}[/tex]
or
the area of the steel rod required = [tex]\frac{\textup{35,000}}{\textup{20,000}}[/tex]
or
the area of the steel rod required = 1.75 in²
Also, Area of rod = [tex]\frac{\pi}{4}d^2[/tex]
where, d is the diameter of the rod
1.75 in² = [tex]\frac{\pi}{4}d^2[/tex]
or
d = 1.49 in
So we can use [tex]1\frac{1}{2}[/tex] in dia rod
An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50 km/h? If the vehicle accelerates to 100 km/h in 15 seconds, what is the change in power required?
Answer:
a)Ek=115759.26J
b)23.15kW
Explanation:
kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation
[tex]Ek=0.5mv^{2}[/tex]
for the first part
m=1200Kg
V=50km/h=13.89m/s
solving for Ek
[tex]Ek=0.5(1200)(13.89)^2[/tex]
Ek=115759.26J
For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.
m=1200kg
V=100km/h=27.78m/s
[tex]Ek=0.5(1200)(27.78)^2=462963J\\[/tex]
taking into account all of the above the following equation is inferred
ΔE=[tex]\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15} =23146.916W=23.15kW[/tex]
To 3 significant digits, what is the temperature of water in degrees C, if its pressure is 350 kPa and the quality is 0.01
Answer:
138.9 °C
Explanation:
The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C
There is 120 lbm. of saturated liquid water in a steel tank at 177 oF. What is the pressure and volume of the tank?
Answer:
P=7.027psi
V=1.9788ft ^3
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
For the first part of this problem, we must find the water saturation pressure at a temperature of 177F
Psat @(177F)=7.027psi
For the second part of this problem, we calculate the specific volume of the water knowing the temperature and that the state is saturated liquid, then multiply by the mass to know the volume
v@177F=0.01649ft^3/lbm
V=mv
V=(0.01649ft^3/lbm)(120lbm)
V=1.9788ft ^3
To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws 12.5 hp for 16 h/day, five days per week. Compute the energy used by the mo- tor for one year. Express the result in ft.lb and W.h
The energy used by the motor for one year is 102,313,596,240 ft.lb/year, expressed in both ft.lb and W.h.
1 horsepower (hp) is approximately equal to 745.7 watts (W).
So,[tex]12.5 hp * 745.7 W/hp[/tex]
= 9312.5 W.
The motor draws 9312.5 watts for 16 hours per day.
So, [tex]9312.5 W * 16 hours/day[/tex]
= 148,200 W.h/day.
Multiply the daily energy by the number of days per year:
The motor operates for 5 days per week, so in one year (considering 52 weeks), it operates for 5 days/week × 52 weeks/year = 260 days/year.
[tex]148,200 W.h/day *260 days/year[/tex]
= 38,532,000 W.h/year.
To express the result in ft.lb, we need to convert the energy from watt-hours to foot-pounds:
1 watt-hour (Wh) is approximately equal to 2655.22 foot-pounds (ft.lb).
So, [tex]38,532,000 W.h/year * 2655.22 ft.lb/W.h[/tex]
≈ 102,313,596,240 ft.lb/year.
In a circular tube the diameter changes abruptly from D1 = 2 m to D2 = 3 m. The flow velocity in the part with smaller diameter is vi = 3 m/s. Determine if for water in the both parts of the tube there is laminar flow or tubulent flow. The kinematic viscosity of water is v= 1.24. 10^-6
Answer:
The flow is turbulent at both the parts of the tube.
Explanation:
Given:
Water is flowing in circular tube.
Inlet diameter is [tex]d_{1}= 2[/tex]m.
Outlet diameter is [tex]d_{2}= 3[/tex]m.
Inlet velocity is [tex]V_{1}= 3[/tex] m/s.
Kinematic viscosity is [tex]\nu =1.24\times 10^{-6}[/tex] m²/s.
Concept:
Apply continuity equation to find the velocity at outlet.
Apply Reynolds number equation for flow condition.
Step1
Apply continuity equation for outlet velocity as follows:
[tex]A_{1}V_{1}=A_{2}V_{2}[/tex]
[tex]\frac{\pi}{4}d^{2}_{1}V_{1}=\frac{\pi}{4}d^{2}_{2}V_{2}[/tex]
Substitute the values in the above equation as follows:
[tex]\frac{\pi}{4}2^{2}\times 3=\frac{\pi}{4}3^{2}V_{2}[/tex]
[tex]2^{2}\times 3=3^{2}V_{2}[/tex]
[tex]V_{2}=\frac{4}{3}[/tex] m/s.
Step2
Apply Reynolds number formula for the flow condition at inlet as follows:
[tex]Re=\frac{v_{1}d_{1}}{\nu }[/tex]
[tex]Re=\frac{2\times 3}{1.24\times 10^{-6}}[/tex]
Re=4838709.677
Apply Reynolds number formula for the flow condition at outlet as follows:
[tex]Re=\frac{v_{2}d_{2}}{\nu }[/tex]
[tex]Re=\frac{\frac{4}{3}\times 3}{1.24\times 10^{-6}}[/tex]
Re=3225806.452
Thus, the Reynolds number is greater than 2000. Hence the flow is turbulent.
Hence, the flow is turbulent at both the parts of the tube.
Starting from position s = 0 m at time t = 0 s, a particle travels on a straight-line path. The particle's velocity is given by the function v = 5 sin (4s2) m/s, where s is in meters and the argument of the sine function is unitless. Find the particle's acceleration when s = 0.25 m and s = 1.25 m.
Answer:
acceleration is 0.2181 m/s²
acceleration is 27.05 m/s²
Explanation:
given data
s = 0
time t = 0
velocity V = 5 sin (4s²) m/s
to find out
particle's acceleration
solution
we have given velocity = 5 sin(4s²)
and we know here that velocity = [tex]\frac{ds}{dt}[/tex]
so acceleration will be a = [tex]\frac{dv}{dt}[/tex]
put here velocity v
acceleration = [tex]\frac{dv}{dt}[/tex]
acceleration = [tex]\frac{d(5sin(4s^4))}{dt}[/tex]
acceleration = 5 cos4s² × 8s × [tex]\frac{ds}{dt}[/tex]
acceleration = 5 cos4s² × 8s × 5 sin4s²
acceleration = 200 s ×cos4s² × sin4s²
put here s = 0.25 and s = 1.25
so
acceleration = 200× 0.25 ×cos(4×0.25²) × sin(4×0.25²)
acceleration = 0.2181 m/s²
acceleration = 200× 1.25 ×cos(4×1.25²) × sin(4×1.25²)
acceleration = 27.05 m/s²
Given 10 parts per billion. What is the concentration
inmg/L?
Answer:
The concentration in mg/L equals 0.01 mg/L
Explanation:
Since the concentration is given as 10 parts per billion
We know that
1 part per billion equals 0.001 mg/L
Thus 10 parts per billion concentration equals
[tex]10\times 0.001mg/L\\\\=0.01mg/L[/tex]
S1.1 The Acre-Foot. Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumped 2.0 in. of rain in 30 minutes on a town of area 26 km2. What volume of water, in acre-feet, fell on the town? S1.2 The Thunderstorm. What mass of water fell on the town in S1.1The Acre-Foot during the thunderstorm? One cubic meter of water has a mass of 103kg.
Answer:
1072 acre foot
1331424000 kg
Explanation:
1 feet has 12 inches, so 2 in is 0.167 feet.
1 km^2 has 1 million m^2.
1 acre is 4074 m^2.
So, 1 km is 247 acres.
Then 26 km^2 is 6422 acres.
So, the volume of water is
6422 * 0.167 = 1072 acre-foot
Since one cubic meter of water has 1000 kg
One inch is 25.4 mm = 0.0254 m
One feet is 12 * 0.0254 = 0.3048 m
An acre-feet has a volume of
4074*0.3048 = 1242 m^3
And that is a mass of water of
1242 * 1000 = 1242000 kg/acre-feet
Therefore the mass of rainwater in the town is of
1072 * 1242000 = 1331424000 kg = 1331424 tons
What does stall mean?
Answer:
Explanation:
In aeronautics stall is the separation of the boundary layer from the wings of plane. This causes loss of lift because the otherwise low pressure area above the wing become filled with turbulent whirls.
Stall occurs when the critical angle of attack of the wing is exceeded.
The highest cutting temperature is located close to the shear zone. a) True b) False
Answer:
a)True
Explanation:
In cutting action,three type of zone are presents
1.Primary zone :
The most of part of energy is converted in to heat.
2.Secondary zone:
Heat generation due to rubbing between tool and chip.This also called deformation zone.
3.Tertiary zone:
Heat generated due to flank of tool and already machined surface.
So the highest temperature is located close to the shear zone.